hdu 2460 Network Tarjan缩点+LCA 解题报告
Problem Description
A network administrator manages a large network. The network consists of N computers and M links between pairs of computers. Any pair of computers are connected directly or indirectly by successive links, so data can be transformed between any two computers. The administrator finds that some links are vital to the network, because failure of any one of them can cause that data can’t be transformed between some computers. He call such a link a bridge. He is planning to add some new links one by one to eliminate all bridges.
You are to help the administrator by reporting the number of bridges in the network after each new link is added.
Input
The input consists of multiple test cases. Each test case starts with a line containing two integers N(1 ≤ N ≤ 100,000) and M(N - 1 ≤ M ≤ 200,000).
Each of the following M lines contains two integers A and B ( 1≤ A ≠ B ≤ N), which indicates a link between computer A and B. Computers are numbered from 1 to N. It is guaranteed that any two computers are connected in the initial network.
The next line contains a single integer Q ( 1 ≤ Q ≤ 1,000), which is the number of new links the administrator plans to add to the network one by one.
The i-th line of the following Q lines contains two integer A and B (1 ≤ A ≠ B ≤ N), which is the i-th added new link connecting computer A and B.
The last test case is followed by a line containing two zeros.
Output
For each test case, print a line containing the test case number( beginning with 1) and Q lines, the i-th of which contains a integer indicating the number of bridges in the network after the first i new links are added. Print a blank line after the output for each test case.
Sample Input
3 2
1 2
2 3
2
1 2
1 3
4 4
1 2
2 1
2 3
1 4
2
1 2
3 4
0 0
Sample Output
Case 1:
1
0
Case 2:
2
0
题意：
给出一个无向图以及Q次询问，每次询问增加一条无向边，要求输出增加这条边后剩余的桥的数目。
思路：
先Tarjan割点，做一次dfs求出所有的桥，当一次询问加入一条边(a,b)之后，会在树上形成一个环，在这个环上的桥都变为非桥，并且这个环肯定经过a和b的LCA，此时我们只需在求LCA的过程中把经过的为桥的树边标记为非桥，同时cnt_bridge再输出即可。
需要注意的是树边的编号是用树边指向的那个节点的编号来表示的，例如树边
代码
最近老是刷这种不清真的题目。我开始怀念板子裸题1A的时候了。。。
#includeint v=G[u][i];
if (!dfn[v])
{
child++;
int lowv=dfs(v,u);
lowu=min(lowu,lowv);
if (lowv>dfn[u])
{
isbridge[v]=1;
cnt_bridge++;
}
}
else if (dfn[v]if (fa<0&&child==1) iscut[u]=0;
return low[u]=lowu;
}
void LCA(int a,int b)
{
while(dfn[a]>dfn[b])
{
if (isbridge[a]) {isbridge[a]=0;cnt_bridge--;}
a=father[a];
}
while(dfn[b]>dfn[a])
{
if (isbridge[b]) {isbridge[b]=0;cnt_bridge--;}
b=father[b];
}
if (a!=b)
{
while(dfn[a]>dfn[b])
{
if (isbridge[a]) {isbridge[a]=0;cnt_bridge--;}
a=father[a];
}
while(dfn[b]>dfn[a])
{
if (isbridge[b]) {isbridge[b]=0;cnt_bridge--;}
b=father[b];
}
}
}
int main()
{
int n,m,Case=1;
while(scanf("%d %d",&n,&m))
{
if (n==0&&m==0) break;
memset(isbridge,0,sizeof(isbridge));
memset(dfn,0,sizeof(dfn));
cnt_bridge=tim=0;
for (int i=0;ifor (int i=0;iint a,b;
scanf("%d %d",&a,&b);
a--;b--;
G[a].push_back(b);
G[b].push_back(a);
}
dfs(0,-1);
scanf("%d",&Q);
printf("Case %d:\n",Case++);
while(Q--)
{
int a,b;
scanf("%d %d",&a,&b);
a--;b--;
LCA(a,b);
printf("%d\n",cnt_bridge);
if (Q==0) printf("\n");
}
}
return 0;
}